equals that of the circular sector ABO with closer and closer approximation. Thus, taking the arcs minutely short, and calling the arc length AB by the letter p, we find

Circular sectorial area = = r.p.

For the complete circular area the length of arc p becomes the complete circumference of the circle or 2πr. Thus the complete circular area is 2.

Here the important point to note is, that the approximation of the sum of the triangular areas to equality with the circular area proceeds pari passu with that of the sum of the tangent lengths to the circular arc length.

60. Constant Gradient. In this integration the factor r is a constant. The variable is p, the arc length measured from some The increment of p is given point on the circular circumference. Sp: that is, Sp may be looked upon as the increase of the arc length p, as the sectorial area is swept out by a radius revolving round the centre O. The integral is the gradually increasing area so swept out. The increment, or differential of this area, is r.dp. Thus r is the differential coefficient of the area with respect to p; or r is the p-gradient of the area. Calling the constant r by the letter k, we may write this

d

dp

(kp)=k; or otherwise

[kdp=

kdp=kp.*

61. Area of Expanding Circle.-A second simple example of integration is that of the area swept through by the circumference of a gradually expanding circle. Let the radius from which the expansion begins be called r1: the area inside this initial circumference is πr12. At any stage of the expansion when the radius has become any size, the area swept through has been (π2 - πr12). This is the indefinite integral; the constant of integration being here — πTM‚2.

As r increases by dr from (r-dr) to (r+r), the area swept out by the circumference is a narrow annular strip whose mean peripheral length is 2πr, and whose radial width is dr, as shown on fig. 10. The area of this annular strip, therefore, equals 2πr.dr, and this is the increment, or differential, of the integral area swept The "differential coefficient with respect to r," or the "r-gradient," of the area is, therefore, 2r. That is, the r-gradient of (πr2+C), where C is any constant, is 2r. Otherwise written,

out.

П

Here is a constant multiplier, and, if any other constant multiplier k were used, the result would be

Written in the inverse manner, the same result appears as

√2krdr =

rdr = kr2+C.*

62. Rectangular Area.-A third simple example is shown in fig. 11. Here a vertical line of length is moved horizontally to the right from an initial position . Its lower end moves along the axis of x. Its upper end moves along a horizontal straight

2

FIG. 11.

line at the height k above this axis. The vertical sweeps out a rectangular area equal to k multiplied by the length of horizontal movement. At any stage x of the movement the area swept out

* See Classified List, III. A. 2.

is k(x-x1). Here the variable is x. Call its increment Sx. While the vertical moves horizontally Sx, the area swept out is a narrow rectangular strip equal to kdx. This is the increment of the integral area. Thus the x-gradient of the area is k. Calling the constant-kx1 by the letter C, this result may be written in the two forms

This result is identical in form with that of § 60, fig. 9, except that in the latter the integration constant is zero and does not appear, and the variable is named p, while x is the name adopted in fig. 11. The choosing of one or other name to indicate the variable is, of course, of no consequence; but taken as two illustrations of purely geometric integration, the two results are of different kinds, although taken as formulas simply they are identical.

63. Triangular Area.-In fig. 12 we have an area swept out by a vertical line as it moves from left to right, and increases in height at a uniform rate during the motion. Its lower end lies always in the axis of x. Its upper

m

1

FIG. 12.

end lies in a straight line whose inclination to the x-axis is called m.

During the movement Sx, from (x − 18x) to (x+8), the mean height of the small strip of area swept out is (k+mx), and its area, therefore, is

(k+mx)dx.

The whole area swept out during the motion from the lower limit x to any upper limit x may be divided into two parts; one rectangular with the base (x-x) and the height (k+ma); the other triangular with the same base and the height m(x-x1).

* See Classified List, III. A. 1.

This whole area is, therefore,

(x − x1) { (k+mx1) +‡m(x − x1)} = k(x − x1) + §m(x2 − x ̧2) .

This may be written in the form

which form means that the function of a standing inside the square bracket is to be calculated for the two values of x, x and x, (the former "indefinite" or not particularised, the latter special), and the latter value of the function subtracted from the former.* Again, it may be written in the other form

This result may be expressed by either of the two following equations:

and

d

dx

{

kx + mx2 + C

}

64. First and Second Powers of Variable.--The last integral may be split into two parts. The first is

which is identical with what is obtained in fig. 11. The second is

[mxdx = mx2 + C2

which is the sweeping out of the triangular area.

65. Integral Momentum.-The following are other easy examples of the first of these two formulas.

The extra momentum acquired by a mass m in the interval between time t1 and time t2, during which its velocity is accelerated at the constant rate g, if its velocity be v1 at time t1, is

Here mg is the acceleration of momentum, or the time-gradient of the momentum.

66. Integral Kinetic Energy.-The simultaneous increase of Kinetic Energy is

= Extra Acquired momentum × Average Velocity during interval.

67. Motion integrated for Velocity and Time.-Again the distance travelled by a train between the times t1 and t2, when running at a constant velocity v, is

Here the velocity v is the time-gradient of the distance travelled, 68. Motion from Acceleration and Time.-Easy examples of the second formula are the following:

If the velocity of a mass be accelerated at the uniform rate g; then, since the velocity at any time t is {g(t − t1)+1}, and since in a small interval of time St, the distance travelled is v.dt, where v is the average velocity during dt, we find the distance travelled in interval (to-t1) to be

{gt — gt1 + v1 } dt = 2 (t2,2 — t12) — gt1(t1⁄2 − t1) + v1(t2 — tı)

= (t2− t1){v1+1g(t2 − t1)}.

If this be multiplied by mg, we get again the increase of kinetic energy as shown above in § 66; so that the increase of kinetic energy equals the uniform acceleration of momentum (mg) multiplied by the distance travelled.

69. Bending Moments.—As another example, take a horizontal beam loaded uniformly with a load w per foot length. If we name by the letter l lengths along the beam from any section where we wish to find the bending moment due to this load; then on any short length dl there is a load w.dl, and the moment of this load upon the given section is wl.dl, where 7 means the length to the

C